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Serga [27]
3 years ago
5

Iron-59 has a half-life of 45.1 days. How old is an iron nail if the Fe-59 content is 25% that of a new sample of iron? Show all

calculations leading to a solution.
Chemistry
1 answer:
alexdok [17]3 years ago
4 0

Answer:  

90.2 da

Explanation:  

The <em>half-life</em> of iron-59 (45.1 da) is the time it takes for half the Fe to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of    Time    Percent       Fraction

<u>half-lives</u>  <u>days</u>   <u>remaining</u>   <u>remaining </u>

      1           45.1         50                 ½

      2          90.2        25                 ¼

      3          135           12.5              ⅛

      4          180            6.25           ¹/₁₆

We see that 25 % remains after two half-lives or 90.2 da.

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14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
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Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
3 years ago
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